Problem: Multiply the following complex numbers: $({i}) \cdot ({1+2i})$
Solution: Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({i}) \cdot ({1+2i}) = $ $ ({0} \cdot {1}) + ({0} \cdot {2}i) + ({1}i \cdot {1}) + ({1}i \cdot {2}i) $ Then simplify the terms: $ (0) + (0i) + (1i) + (2 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ 0 + (0 + 1)i + 2i^2 $ After we plug in $i^2 = -1$ , the result becomes $ 0 + (0 + 1)i - 2 $ The result is simplified: $ (0 - 2) + (1i) = -2+i $